Is the tangent function (like in trig) and tangent lines the same?

So, a 45 degree angle in the unit circle has a tan value of 1. Does that mean the slope of a tangent line from that point is also 1? Or is something different entirely?

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Have a look at this drawing from Wikipedia: Unit Circle Definitions of Trigonometric Functions.

When viewed this way, the tangent function actually represents the slope of a line perpendicular to the tangent line of that point (i.e. the slope of the radius that touches the angle point).

However, you can actually see that the "tangent line", consisting the values of the tangents, is the actual tangent line of the circle at the point from which the angles are measured, and I would guess that this is the source of the name.

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The latin tangere means "to touch". For the calculus definition at least, this probably relevant. – Robert Mastragostino Dec 27 '12 at 3:51
Interesting! Never thought of that... like "tangible".Here is an (unsourced) description of origins of the names of the trigonometric functions. – Alfonso Fernandez Dec 27 '12 at 4:52

Yes and no, resp.: yes, any line in the plane that forms an angle of $\,45^\circ\,$ with the positive direction of the $\,x-$axis has a slope of $\,\tan 45^\circ=1\,$, and no: it isn't something different.

It is not completely clear though what you mean by "tangent line"...perhaps you meant "tangent line at some point on the graph of a (derivable) function"?

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I don't know to reply directly to you, Don. Anyway, what I mean is this. tan(60 degrees) is sqrt(3). What's that mean though? Does it mean the slope of a tangent line at the point, let's use the unit circle, (1/2, sqrt(3)/2) is sqrt(3)? My confusion is what that sqrt(3) is meant to represent. – user54350 Dec 27 '12 at 2:44
Yes, of course...in fact, that is the direct definion from trigonometry and a straight angle triangle: $$\tan 60^\circ=\frac{\text{opposite leg}}{\text{adjacent leg}}=\frac{\frac{\sqrt 3}{2}}{\frac{1}{2}}=\sqrt 3$$ The $\,\sqrt 3\,$ represents the ration of the lengths of the two legs in that triangle. – DonAntonio Dec 27 '12 at 2:46
That's what I thought. Thank you, but I have a follow up question. If tan if y/x (unit circle), then tan at 0 degrees is 0/1, which is zero. But wouldn't a tangent line from that point just be a vertical line, which has an undefined slope? Similarly, at a 90 degree angle tan = 1/0, which is undefined, but a tangent line from that point would be horizontal, no? Here's an image that I hope will show the contradiction that I'm trying to get cleared up. i.imgur.com/eNm0j.jpg – user54350 Dec 27 '12 at 2:57
Remember that a horizontal line (=parallel to the $\,-$axis) is the one that has zero slope = $\,\tan 0\,$ , as the difference between the $\,y-$coordinates is zero. A vertical line has no defined slope, though later (much later) one could assign it, under certain conditions, an infinite slope. BTW, your drawing shows you're confusing the definition of tangent: it is "difference of y-coordinates divided by difference of x-coordinates", as long as the later is non-zero – DonAntonio Dec 27 '12 at 3:01
How would you draw that on the unit circle? – user54350 Dec 27 '12 at 3:08